Question

Help me solving this problem -

The probability that you manage to get ready on time at the start of a given day is 0.7 and when you are ready on time you can then take the university bus. Otherwise, you have to take a public/local bus to get to the campus. The university bus is equally likely to take 41, 42, ., ог 50 minutes to reach the campus, whereas the local/public bus is equally likely to take 46, 47, or 55 minutes (integral measures in both cases). What is the expected time for you to get to the campus on any given day? What is the probability that you'd manage to get to the campus within 45 minutes on any given day?

Answer 1

The expected time for the student to get to campus is 46.94 minutes, and the probability of getting to campus within 45 minutes is 35%.

Step-by-step explanation:

Expected Time and Probability Calculation

To solve this problem, we must calculate the expected commute time on a given day, and the probability of getting to the campus within 45 minutes. Since the student manages to get ready on time with a probability of 0.7, they will take the university bus 70% of the time. The university bus takes between 41 and 50 minutes with equal probability, so we calculate the average time by summing these times and dividing by the number of values (10). For the local/public bus, which the student takes when not ready on time (30% probability), we similarly calculate the average time over the three possible values.

Expected Time Calculation:

University bus average time = (41+42+...+50) / 10 = 45.5 minutes

Local bus average time = (46+47+55) / 3 = 49.3 minutes (rounded to one decimal place)

Overall expected time = (0.7 * 45.5) + (0.3 * 49.3) = 46.94 minutes

Probability of getting to campus within 45 minutes:

With university bus: P(41 or 42 or 43 or 44 or 45) = 5/10 = 0.5

With local bus: P(46 or 47 or 55) < 45 minutes = 0, since none of these times are less than 45 minutes.

Overall probability = (0.7 * 0.5) + (0.3 * 0) = 0.35 or 35%